Question 724435
find the vertices and foci of the hyperbola 
4x²-y²-48x-4y+40=0
4x^2-48x-y^2-4y=-40
complete the square:
4(x^2-12x+36)-(y^2+4y+4)=-40+144+4
4(x-6)^2-(y+2)^2=108
{{{(x-6)^2/27-(y+2)^2/108=1}}}
This is an equation of a hyperbola with horizontal transverse axis
Its standard form: {{{(x-h)^2/a^2-(y-k)^2/b^2=1}}}, (h,k)=(x,y)coordinates of the center.
For given equation: 
center: (6,-2)
a^2=27
a=√27≈5.2
b^2=108
c^2=a^2+b^2=27+108=135
c=√135≈11.6
vertices: (6±a,-2)=(6±5.2,-2)=(.8,-2) and (11.2,-2)
foci: (6±c,-2)=(6±11.6,-2)=(-5.6,-2) and (17.6,-2)